Activity selection problem explained

The activity selection problem is a combinatorial optimization problem concerning the selection of non-conflicting activities to perform within a given time frame, given a set of activities each marked by a start time (si) and finish time (fi). The problem is to select the maximum number of activities that can be performed by a single person or machine, assuming that a person can only work on a single activity at a time. The activity selection problem is also known as the Interval scheduling maximization problem (ISMP), which is a special type of the more general Interval Scheduling problem.

A classic application of this problem is in scheduling a room for multiple competing events, each having its own time requirements (start and end time), and many more arise within the framework of operations research.

Formal definition

Assume there exist n activities with each of them being represented by a start time si and finish time fi. Two activities i and j are said to be non-conflicting if sifj or sjfi. The activity selection problem consists in finding the maximal solution set (S) of non-conflicting activities, or more precisely there must exist no solution set S' such that |S'| > |S| in the case that multiple maximal solutions have equal sizes.

Optimal solution

The activity selection problem is notable in that using a greedy algorithm to find a solution will always result in an optimal solution. A pseudocode sketch of the iterative version of the algorithm and a proof of the optimality of its result are included below.

Algorithm

Greedy-Iterative-Activity-Selector(A, s, f):

Sort A by finish times stored in f S = k = 1 n = A.length for i = 2 to n: if s[i] ≥ f[k]: S = S U k = i return S

Explanation

Line 1: This algorithm is called Greedy-Iterative-Activity-Selector, because it is first of all a greedy algorithm, and then it is iterative. There's also a recursive version of this greedy algorithm.

A

is an array containing the activities.

s

is an array containing the start times of the activities in

A

.

f

is an array containing the finish times of the activities in

A

.

Note that these arrays are indexed starting from 1 up to the length of the corresponding array.

Line 3: Sorts in increasing order of finish times the array of activities

A

by using the finish times stored in the array

f

. This operation can be done in

O(nlogn)

time, using for example merge sort, heap sort, or quick sort algorithms.

Line 4: Creates a set

S

to store the selected activities, and initialises it with the activity

A[1]

that has the earliest finish time.

Line 5: Creates a variable

k

that keeps track of the index of the last selected activity.

Line 9: Starts iterating from the second element of that array

A

up to its last element.

Lines 10,11: If the start time

s[i]

of the

ith

activity (

A[i]

) is greater or equal to the finish time

f[k]

of the last selected activity (

A[k]

), then

A[i]

is compatible to the selected activities in the set

S

, and thus it can be added to

S

.

Line 12: The index of the last selected activity is updated to the just added activity

A[i]

.

Proof of optimality

Let

S=\{1,2,\ldots,n\}

be the set of activities ordered by finish time. Assume that

A\subseteqS

is an optimal solution, also ordered by finish time; and that the index of the first activity in A is

k1

, i.e., this optimal solution does not start with the greedy choice. We will show that

B=(A\setminus\{k\})\cup\{1\}

, which begins with the greedy choice (activity 1), is another optimal solution. Since

f1\leqfk

, and the activities in A are disjoint by definition, the activities in B are also disjoint. Since B has the same number of activities as A, that is,

|A|=|B|

, B is also optimal.

Once the greedy choice is made, the problem reduces to finding an optimal solution for the subproblem. If A is an optimal solution to the original problem S containing the greedy choice, then

A\prime=A\setminus\{1\}

is an optimal solution to the activity-selection problem

S'=\{i\inS:si\geqf1\}

.

Why? If this were not the case, pick a solution B′ to S′ with more activities than A′ containing the greedy choice for S′. Then, adding 1 to B′ would yield a feasible solution B to S with more activities than A, contradicting the optimality.

Weighted activity selection problem

The generalized version of the activity selection problem involves selecting an optimal set of non-overlapping activities such that the total weight is maximized. Unlike the unweighted version, there is no greedy solution to the weighted activity selection problem. However, a dynamic programming solution can readily be formed using the following approach:[1]

Consider an optimal solution containing activity . We now have non-overlapping activities on the left and right of . We can recursively find solutions for these two sets because of optimal sub-structure. As we don't know, we can try each of the activities. This approach leads to an

O(n3)

solution. This can be optimized further considering that for each set of activities in

(i,j)

, we can find the optimal solution if we had known the solution for

(i,t)

, where is the last non-overlapping interval with in

(i,j)

. This yields an

O(n2)

solution. This can be further optimized considering the fact that we do not need to consider all ranges

(i,j)

but instead just

(1,j)

. The following algorithm thus yields an

O(nlogn)

solution:

Weighted-Activity-Selection(S): // S = list of activities

sort S by finish time opt[0] = 0 // opt[j] represents optimal solution (sum of weights of selected activities) for S[1,2..,j] for i = 1 to n: t = binary search to find activity with finish time <= start time for i // if there are more than one such activities, choose the one with last finish time opt[i] = MAX(opt[i-1], opt[t] + w(i)) return opt[n]

External links

Notes and References

  1. http://www.cs.princeton.edu/~wayne/cs423/lectures/dynamic-programming-4up.pdf Dynamic Programming with introduction to Weighted Activity Selection